Some theorem regarding rational numbers

If y>x where x and y are both elements of the reals, but x is also irrational, I must prove that there is a rational number z such that x<z<y. I can only show this is true when x is rational. How do you add something to an irrational number to make it rational?

If y>x where x and y are both elements of the reals, but x is also irrational, I must prove that there is a rational number z such that x<z<y. I can only show this is true when x is rational. How do you add something to an irrational number to make it rational?

Then [tex]x+r[/tex] is rational because it will end in [tex]\bar{9}[/tex], and
[tex]0 \leq r < \epsilon [/tex]. so
[tex]x \leq x + r < x + \epsilon \rightarrow x \leq x+r < y[/tex]
Which is what you wanted to prove.
You can make the [tex]\leq[/tex] strict if you note that x is irrational, and [tex]x+r[/tex] is rational, thus they cannot be equal.